Key Concepts and Formulas

• Area of a Triangle (Coordinate Geometry): The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by: $Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
• Inscribed Parallelogram Property: The maximum area of a parallelogram AFDE inscribed in $\triangle ABC$ such that A is a vertex of the parallelogram and D, E, and F lie on BC, CA, and AB respectively, is half the area of $\triangle ABC$.

Step-by-Step Solution

Step 1: Calculate the area of $\triangle ABC$ We are given the coordinates of the vertices A(4, -2), B(1, 1), and C(9, -3). We will use the formula for the area of a triangle given its vertices. $Area = \frac{1}{2} |4(1 - (-3)) + 1(-3 - (-2)) + 9(-2 - 1)|$ $Area = \frac{1}{2} |4(4) + 1(-1) + 9(-3)|$ $Area = \frac{1}{2} |16 - 1 - 27|$ $Area = \frac{1}{2} |-12|$ $Area = \frac{1}{2} (12)$ $Area = 6$ Thus, the area of $\triangle ABC$ is 6 square units.

Step 2: Determine the maximum area of parallelogram AFDE According to the inscribed parallelogram property, the maximum area of parallelogram AFDE is half the area of $\triangle ABC$. Therefore, $Area_{AFDE(max)} = \frac{1}{2} \times Area_{\triangle ABC}$ $Area_{AFDE(max)} = \frac{1}{2} \times 6$ $Area_{AFDE(max)} = 3$ The maximum area of the parallelogram AFDE is 3 square units.

Step 3: Compare with Options Looking at the provided options, none of them match the calculated answer of 3. The "Correct Answer" listed is 1, which is incorrect. Since we have rigorously derived the correct answer as 3, there seems to be an error either in the options or in the marked correct answer.

Common Mistakes & Tips

• Sign Errors: Be careful with the signs when applying the area formula. A single sign error can lead to an incorrect area.
• Formula Confusion: Make sure you are using the correct formula for the area of a triangle given the coordinates of its vertices.
• Inscribed Parallelogram Property: Remember that the maximum area of the inscribed parallelogram is half the area of the triangle.

Summary

We calculated the area of $\triangle ABC$ using the coordinate geometry formula and found it to be 6 square units. Using the property that the maximum area of the inscribed parallelogram AFDE is half the area of $\triangle ABC$, we determined that the maximum area of parallelogram AFDE is 3 square units. The "Correct Answer" given is incorrect.

Final Answer

The final answer is \boxed{3}. There appears to be an error in the problem statement, as the correct answer should be 3, not 1.